Class Notes (1,100,000)
CA (630,000)
U of G (30,000)
BIOL (2,000)
BIOL 1090 (600)
Lecture 6

BIOL 1090 Lecture Notes - Lecture 6: Taa Language, Albinism

Course Code
BIOL 1090
Andrew Bendall

This preview shows half of the first page. to view the full 2 pages of the document.
BIOL1090 Lecture 6 27 Jan 2016
Pedigrees and Human Genetics
Heterozygote can have the same phenotype as a homozygous dominant
o Can use a test cross to figure out genotype
o For an accurate test cross, the organism must be crossed with a homozygous
recessive individual
Show relationship between individuals in a family
o Phenotypes are displayed on pedigree
Circles = female; squares = male
Coloured circle/square = individuals with the tracked trait
Traits are likely dominant if:
Every affected individual has at least one affected parent
The trait is manifested in at least one individual in every generation once the trait
The spontaneous appearance of a dominant allele is extremely rare
Traits are likely recessive if:
The trait suddenly appears in the pedigree
The trait skips a generation
In the absence of evidence to the contrary, assume that unrelated individuals marrying
into the family do not carry the recessive allele
Since human families are relatively small, the phenotypic ratios among offspring often
deviate significantly from Mendelian expectations
Albinism Case Study:
R’s likelihood of being heterozygous is 2/3 instead of 2/4 because we know that he is not
homozygous recessive (showing the trait) because the pedigree shows he doesn’t display
it, therefore we take it out of our calculation
Calculate the risk that child T has albinism:
o PT(aa) = PR(Aa) x PA(a) x PS(a)
o = (2/3) x (1/2) x (1)
o = 2/6
o = 1/3
find more resources at
find more resources at
You're Reading a Preview

Unlock to view full version